Routh-Hurwitz Criterion: Special Cases

Zero only in the first column

When forming the Routh table, replace the

zero with a small number ε and evaluate thefist column for positive or negative values of ε.

Problem: Determine the stability of the closed-

loop transfer function:

T (s) =10

s5+2s4+3s3+6s2+5s+3(1)

s5 1 3 5

s4 2 6 3

s3−

1 32 6

2=6 0→ ε

−1 52 3

2= 7

2

−1 02 0

2= 0

s2

−2 6ε 7

2ε

= 6ε−7ε

−2 3ε 0

ε= 3

−2 0ε 0

ε= 0

s1

−ε 7

26ε−7

ε3

6ε−7ε

= 42ε−49−6ε2

12ε−14

−ε 0

6ε−7ε

06ε−7

ε

= 0

−ε 0

6ε−7ε

06ε−7

ε

= 0

s0 3 0 0

1

Label First Column ε = + ε = −

s5 1 + +

s4 2 + +

s3 ε + −

s2 6ε−7ε − +

s1 42ε−49−6ε2

12ε−14 + +

s0 3 + +

The changes in sign occur in both instances so

it doesn’t matter which we choose. they both

indicate the system is unstable with two poles

in the rhp.

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Another, less computationally expensive method

to use when a zero occurs in the first column

is to create the Routh table using the polyno-

mial that has the reciprocal roots of the origi-

nal polynomial. i.e.

T (s) =10

s5+2s4+3s3+6s2+5s+3(2)

by replacing s with 1/s.

D(s) = 3s5+5s4+6s3+3s2+2s+1 (3)

s5 3 6 2

s4 5 3 1

s3 4.2 1.4 0

s2 1.33 1 0

s1 −1.75 0 0

s0 1 0 0

since there are two sign changes, the system is unstable and has

two right half plane (rhp) poles. This is the same as the result

obtained by using ε for the original polynomial.

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Entire row is zero

Problem: Determine the number of right hand

plane poles in the closed-loop transfer func-

tion:

T (s) =10

s5+7s4+6s3+42s2+8s+56(4)

s5 1 6 8

s4 6 7→ 1 6 42→ 6 6 56→ 8

s3 0→ 0→ 0

s2

s1

s0

Now we are faced with the problem of zeros in

the third row.

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1. Form a new polynomial using the entries inthe row above zeros. The polynomial will startwith power of s in that row, and continue byskipping every other power of s, i.e.

P (s) = s4+6s2+8 (5)

2. Next we differentiate the polynomial withrespect to s and obtain

dP (s)

ds= 4s3+12s+0 (6)

3. Finally the row with all zeros in the Routhtable is replaced with the coefficients in Eq.(6),and continue the table.

s5 1 6 8

s4 6 7→ 1 6 42→ 6 6 56→ 8

s3 6 0→6 4 → 1 6 0→6 12→ 3 6 0→ 0

s2 3 8 0

s1 13

0 0

s0 8 0 0

We see no sign changes hence no rhp poles.

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• Why does an entire row of zeros occur?When a purely odd or even polynomial is afactor of the original polynomial. (s4+6s2+8is an even polynomial as it only has even powerof s.)

σ

j ω

j ωωj

σ

quadrantal and symmetrical

imaginarysymmetrical and symmetrical and real

σ

• Some polynomial only have roots symmetri-cal about the origin.

• Routh table from the even polynomial (s4 →

s0) is a test of the even polynomial.

• The rows of zeros indicates the possilbilityof jω roots.

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Look at a larger example:

T (s) =20

s8 + s7 + 12s6 + 22s5 + 39s4 + 59s3 + 48s2 + 38s + 20(7)

s8 1 12 39 48 20

s7 1 22 59 38 0

s6 − 6 10→ −1 − 6 20→ −2 6 10→ 1 6 20→ 2 0

s5 6 20→ 1 6 60→ 3 6 40→ 2 0 0

s4 1 3 2 0 0

s3 6 0→6 4 → 2 6 0→6 6→ 3 6 0→ 0 0 0

s2 32

2 0 0 0

s1 13

0 0 0 0

s0 4 0 0 0 0

1. Rows of zeros of s3, Form a new polynomial

using the entries in the row above zeros, i.e.

s4.

P (s) = s4+3s2+2 (8)

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2. Next we differentiate the polynomial with

respect to s and obtain

dP (s)

ds= 4s3+6s+0 (9)

3. Finally the row with all zeros in the Routh

table is replaced with the coefficients in Eq.(9),

and continue the table.

• As the entries from s4 to s0 are looking at

the even polynomials and there are no sign

changes. So all four poles must exits on the

jω axis.

• There are two sign changes from s8 to s4, so

there are 2 right hand plane poles and 2 left

hand plane poles.

• Final results: 2rhp, 2lhp, 4 jω poles.

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